HDOJ 题目3555 Bomb(数位DP)
2024-08-24 08:26:24
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 10419 Accepted Submission(s): 3673
Problem Description
The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would
add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?
add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?
Input
The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.
The input terminates by end of file marker.
The input terminates by end of file marker.
Output
For each test case, output an integer indicating the final points of the power.
Sample Input
3
1
50
500
Sample Output
0
1
15HintFrom 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499",
so the answer is 15.
Author
fatboy_cw@WHU
Source
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zhouzeyong | We have carefully selected several similar problems for you:
pid=3554" style="color:rgb(26,92,200); text-decoration:none">3554
3556 3557 3558 3559这样的比递推的那种写法慢了些啊
ac代码
#include<stdio.h>
#include<string.h>
int bit[20];
__int64 dp[20][10][2];
__int64 dfs(int pos,int pre,int isture,int limit)
{
if(pos<0)
{
return isture;
}
if(!limit&&dp[pos][pre][isture]!=-1)
{
return dp[pos][pre][isture];
}
int last=limit? bit[pos]:9;
__int64 ans=0;
for(int i=0;i<=last;i++)
{
ans+=dfs(pos-1,i,isture||(pre==4&&i==9),limit&&(i==last));
}
if(!limit)
{
dp[pos][pre][isture]=ans;
}
return ans;
}
__int64 solve(__int64 n)
{
int len=0;
while(n)
{
bit[len++]=n%10;
n/=10;
}
return dfs(len-1,0,0,1);
}
int main()
{
int n;
memset(dp,-1,sizeof(dp));
int t;
scanf("%d",&t);
while(t--)
{
__int64 n;
scanf("%I64d",&n);
printf("%I64d\n",solve(n));
}
}
人家的代码http://blog.csdn.net/scf0920/article/details/42870573
#include <iostream>
#include <string.h>
#include <math.h>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <map>
#include <set>
#include <stdio.h>
using namespace std;
#define LL __int64
#define pi acos(-1.0)
const int mod=100000000;
const int INF=0x3f3f3f3f;
const double eqs=1e-8;
LL dp[21][11], c[21];
LL dfs(int cnt, int pre, int maxd, int zero)
{
if(cnt==-1) return 1;
if(maxd&&zero&&dp[cnt][pre]!=-1) return dp[cnt][pre];
int i, r;
LL ans=0;
r=maxd==0? c[cnt]:9;
for(i=0;i<=r;i++){
if(!zero||!(pre==4&&i==9)){
ans+=dfs(cnt-1,i,maxd||i<r,zero||i);
}
}
if(maxd&&zero) dp[cnt][pre]=ans;
return ans;
}
LL Cal(LL x)
{
int i, cnt=0;
while(x){
c[cnt++]=x%10;
x/=10;
}
return dfs(cnt-1,-1,0,0);
}
int main()
{
int t;
LL n;
scanf("%d",&t);
memset(dp,-1,sizeof(dp));
while(t--){
scanf("%I64d",&n);
printf("%I64d\n",n+1-Cal(n));
}
return 0;
}
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